The quantities
$$\boldsymbol{m}=\boldsymbol{r}\times\boldsymbol{p}, \qquad \boldsymbol{W}=\boldsymbol{p}\times\boldsymbol{m}+ \mu\alpha\frac{\boldsymbol{r}}{|\boldsymbol{r}|}$$
are constants of motion, as is well known.

It is stated then that the flows generated by the functions $m_i$ and $W_i,\ i=1,2,3$ are canonical transformations.

I don't understand is what is meant by this statement: I mean, I know what a canonical transformation is, but I would appreciate some explanation or reference about this precise statement.

Thanks in advance for the help, and I hope this is formulated in compliance with the rules of this community.

1 Answer
1

The flow generated by a function $f$ on the phase space $T^*M$ (whose coordinate functions are $\mathbf{q}$ and $\mathbf{p}$) is a set of new coordinate functions $\mathbf{Q}(\mathbf{q}, \mathbf{p}, t)$ and $\mathbf{P}(\mathbf{q}, \mathbf{p}, t)$, (parameterized by a parameter $t$ (time)) satisfying the first order differential equations (flow equations):

It is very well known the flow of any component of $\mathbf{m}$ or $\mathbf{W}$ generates a canonical transformation, i.e., for each time $t$:

$\{Q_i, Q_j\} = 0$

$\{Q_i, P_j\} = \delta_{ij}$

$\{P_i, P_j\} = 0$.

The reason is that the each component serves as a generating function of canonical transformations (in the sense of Goldstein (Classical Mechanics)).

This statement can be cast in a more modern language of symplectic geometry: The Hamiltonian vector fields corresponding to each component automatically preserve the symplectic form thus the transformations are canonical.

Of course, it is not very hard as an exercise to integrate the flow equations exactly and verify the canonical commutation relations.